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I am reading Peskin & Schroeder's book on Chapter 2. I have a question about how to get the term $\frac{1}{-2E(\vec{p})} e^{-ip(x-y)}$. The original equation for propagator is

$$ \langle 0 | [\phi(x), \phi(y)] | 0 \rangle = \int \frac{d^{3}p}{(2\pi)^3} \frac{1}{2E(\vec{p})} [ e^{-ip(x-y)} - e^{ip(x-y)} ].\tag{2.54} $$ We separate this equation, namely, $$ \langle 0 | [\phi(x), \phi(y)] | 0 \rangle = \int \frac{d^{3}p}{(2\pi)^3} \frac{1}{2E(\vec{p})} e^{-ip(x-y)} + \int \frac{d^{3}p}{(2\pi)^3} \frac{1}{-2E(\vec{p})} e^{ip(x-y)} $$ Peskin & Schroeder's Book says that the energy $p^{0}=-E(\vec{p})$ in the second term is less than 0. I tried to expand $e^{ip(x-y)}$ in the following:

$$ e^{ip(x-y)} = e^{i(p^{\mu}(x-y)_{\mu})} = e^{i(Et-\vec{p}\cdot(\vec{x}-\vec{y}))} = e^{i[-(-Et)-\vec{p}\cdot(\vec{x}-\vec{y})]}=e^{-i[p^{0}t+\vec{p}\cdot(\vec{x}-\vec{y})]} $$ However, it seems that I can not write $e^{-i[p^{0}t+\vec{p}\cdot(\vec{x}-\vec{y})]}$ as $e^{-ip(x-y)}$, where $p^{0}=-E(\vec{p})$. Where is the problem?

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    $\begingroup$ In the second term $\vec p$ is a dummy variable of integration, so you can change variables from $\vec p$ to $-\vec p$. It doesn't affect the measure and it doesn't affect $E(\vec p)=|\vec p|^2 + m^2$. $\endgroup$
    – hft
    Jul 27, 2022 at 4:24
  • $\begingroup$ @hft Hi, Thanks. I understand. $\endgroup$
    – quantum
    Jul 27, 2022 at 7:54

1 Answer 1

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Yeah, basically as hft said in his comment, the main point is the integrals are only in the space/momentum components and not in time/energy components.

And because you are integrating over all momenta $\vec{p}$, for every $\vec{x}-\vec{y}$ you pass, you will also pass through a $\vec{y}-\vec{x}$. Meaning that effectively, and only inside the integral: \begin{equation} \vec{x}-\vec{y}=\vec{y}-\vec{x} \end{equation} and because this only happens in space components the time component of $p$ which is $E$ will gain a minus sign, telling you that you are dealing with "negative energy" particles (antiparticles):

$$ \langle 0 | [\phi(x), \phi(y)] | 0 \rangle = \int \frac{d^{3}p}{(2\pi)^3} \frac{1}{2E(\vec{p})} e^{-ip(x-y)} \Big|_{p_0=E} + \int \frac{d^{3}p}{(2\pi)^3} \frac{1}{2(-E(\vec{p}))} e^{-ip(x-y)} \Big|_{p_0=-E} $$

Notice you could also have absorbed the minus in the time component of $x$ instead, telling you that those antiparticles can be thought as normal particles kind of travelling back in time: $$ \langle 0 | [\phi(x), \phi(y)] | 0 \rangle = \int \frac{d^{3}p}{(2\pi)^3} \frac{1}{2E(\vec{p})} e^{-ip(x-y)} \Big|_{p_0=E, \ t_+} - \int \frac{d^{3}p}{(2\pi)^3} \frac{1}{2E(\vec{p})} e^{-ip(x-y)} \Big|_{p_0=E, \ t_-} $$ (This last paragraph has to be taken with caution, since in reality is more complex than that)

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